SOLUTION: Find the values of a and b so that the function f(x)=(1/3)x^3 + ax^2 +bx will have a local extrema at (3,1).

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Question 1106177: Find the values of a and b so that the function f(x)=(1/3)x^3 + ax^2 +bx will have a local extrema at (3,1).
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


We know that...

(a) f(3) = 1:
%281%2F3%29%283%5E3%29%2Ba%283%5E2%29%2Bb%283%29+=+1
9%2B9a%2B3b+=+1
9a%2B3b+=+-8 (1)

and

(b) f'(3) = 0:
f'(x) = x^2+2ax+b
9%2B6a%2Bb+=+0
6a%2Bb+=+-9 (2)

Solve the pair of equations (1) and (2).

The numbers turn out to be ugly fractions, so solve for both a and b using elimination.

solve for a (eliminate b):
9a%2B3b+=+-8
18a%2B3b+=+-27
9a+=+-19
a+=+-19%2F9
solve for b (eliminate a):
18a%2B6b+=+-16
18a%2B3b+=+-27
3b+=+11
b+=+11%2F3

The required function is f%28x%29+=+%281%2F3%29x%5E3-%2819%2F9%29x%5E2%2B%2811%2F3%29x

The graph shows a relative minimum at (3,1):